Given the code for the recursive algorithm that sums the sequence of numbers from one to one hundred: def sum(n): if n < 1: return 0 else: return n + sum(n - 1) result = sum(100) print(result) How could the code be altered to print the sum of the first 10 numbers in the sequence 3n−1? (1 point) Responses def sum(n): if n <= 0: return 0 else: return n + sum(3 * n - 1) result = sum(10) print(result) def sum(n): if n <= 0: return 0 else: return n + sum(3 * n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * (n - 1) + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * (n - 1) + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(100) print(result)
def sum(n):
if n == 0:
return 0
else:
return (3 * n) - 1 + sum(n - 1)
result = sum(10)
print(result)
The code can be altered to print the sum of the first 10 numbers in the sequence 3n-1 by using the following code:
def sum(n):
if n <= 0:
return 0
else:
return 3 * n - 1 + sum(n - 1)
result = sum(10)
print(result)
The correct code to print the sum of the first 10 numbers in the sequence 3n−1 would be:
def sum(n):
if n == 0:
return 0
else:
return 3 * n - 1 + sum(n - 1)
result = sum(10)
print(result)
Explanation:
To find the sum of the first 10 numbers in the sequence 3n−1, we need to modify the recursive algorithm as follows:
1. The base case n == 0 is used to stop the recursion when n becomes 0.
2. In each recursive call, we calculate the sum of the current number (3 * n - 1) and the sum of the previous numbers (sum(n - 1)).
3. We pass n - 1 in each recursive call to move towards the base case by reducing the value of n.
The result is then printed using the print() function.